GEOMETRIC CONSTRUCTIONS AND ALGEBRAIC FIELD EXTENSIONS JENNY WANG Abstract. In this paper, we study field extensions obtained by polynomial rings and maximal ideals in order to determine whether solutions exist to three ancient Greek construction problems: squaring the circle, doubling the cube, and trisecting an angle. Contents 1. Introduction 1 2. Principal Ideal Domains and Polynomial Division 2 3. Field Extensions 3 4. Algebraic Extensions 6 5. Impossibility of Geometric Constructions 8 5.1. Squaring the Circle 10 5.2. Doubling the Cube 11 5.3. Trisecting an Angle 11 6. Acknowledgments 12 References 12 1. Introduction Much of Ancient Greek mathematics was based in geometry. One particular point of interest was determining which geometric elements could be constructed using only an unmarked straightedge and a compass. It is a quick classroom ex- ercise, for example, to construct congruent circles and perpendicular lines. The constructibility of some other elements, however, is less immediate. Can we con- struct a square with the same area as any given circle? Given a cube, can we construct a second cube with twice the volume? Can we trisect any given angle? To answer these questions, we turn to the study of algebraic structures. Given the geometric nature of the Greek construction problems, it is under- standable if the motivation for studying polynomial rings and field extensions is not entirely apparent. The aim of this paper, however, is to build a fundamental understanding of polynomial rings, maximal ideals, and algebraic extensions in or- der to determine the possibility (or impossibility) of certain constructions without having to explicitly construct the elements themselves. We begin by proving some relevant results about principal ideal domains and polynomial division. In the next few sections, we will determine how to construct 1 2 JENNY WANG and classify field extensions before finally revisiting the geometric construction problems. It should be noted that discussion of rings in this paper will be restricted to commutative rings. Imposing this condition does not detract from the validity of the results eventually obtained and simplifies notation for the reader. 2. Principal Ideal Domains and Polynomial Division We begin with some important properties of polynomial rings and integral do- mains. The reader is assumed to be familiar with the conventional notion of long division of polynomials. Given any polynomial as our dividend, we know how to divide it by a non-zero polynomial to get a quotient and remainder. With a restric- tion only on the degree of the remainder, the Division Algorithm ensures both the existence and uniqueness of the quotient and remainder. We can thus formalize the basic polynomial division with the following theorem: Theorem 2.1 (Division Algorithm). Let F be a field and f and g polynomials in F [x], where g is non-zero. Then there exist unique q; r also in F [x] such that (1) f(x) = q(x)g(x) + r(x) and (2) deg(r) < deg(g) This theorem will prove to be crucial in proving many useful results about princi- pal ideal domains and field extensions. Suppose, for example, we are given any two polynomials f and g in a polynomial ring F [x], where g is non-zero. The Division Algorithm guarantees the existence of two more polynomials q and r so that f can be written as the product of q and g, plus a remainder r. In the special case that the remainder has degree zero, then we say that f is a multiple of g. This basic idea is the conceptual foundation of our discussion of principal ideal domains, as principal ideals in F [x] consist of nothing more than multiples of a given polynomial. We claim that for any field F , the polynomial ring F [x] is a principal ideal domain - meaning every ideal in the ring F [x] is a principal ideal. The proof relies on the existence of a division algorithm in F [x]. Theorem 2.2. Let F be a field. For any ideal I in F [x], I = (a(x)) for some a(x) 2 F [x] Proof. If I = f0g, we have that I = (0) so a(x) is the zero polynomial and I = (a(x)). Now consider when I 6= f0g. We claim that I is generated by a polynomial of minimal degree in the ideal, call it a(x). Note that the existence of a polynomial with minimal degree is guaranteed by the Well-Ordering Principle. Now, because I 6= f0g, a(x) is not the zero polynomial. By Theorem 2.1, for any f(x) 2 I, there exist unique q(x); r(x) 2 F [x] such that f(x) = q(x)a(x) + r(x) where deg(r(x)) < deg(a(x)). Rearranging, we get r(x) = f(x) − q(x)a(x) As f(x) and a(x) are both in the ideal, it follows that r(x) 2 I as well. However, by the condition on degree of r by the division algorithm and the minimality of the degree of a(x), it must be that r(x) is the zero polynomial. Hence for any f 2 I, GEOMETRIC CONSTRUCTIONS AND ALGEBRAIC FIELD EXTENSIONS 3 f(x) = q(x)a(x) for some q(x) 2 F [x]. Therefore I = (a(x)) for any ideal I in F [x]. Once it is established that every ideal in a given ring is generated by a single element, there arises an important correspondence between prime and maximal ideals. Given a prime ideal (p) in a principal ideal domain, we will show there are no proper ideals that contain (p) except (p) itself - meaning it is by definition a maximal ideal. Theorem 2.3. In a principal ideal domain, every prime ideal is maximal. Proof. Consider a prime ideal I of a ring R. If R is a principal ideal domain, then I = (p) for some p 2 R. In order to show that (p) is maximal, it is sufficient to show that any ideal containing (p) is either (p) itself or the entire ring R. So let (m) be an ideal containing (p). Then p = m · r for some r 2 R. As (p) is prime, either m 2 (p) or r 2 (p). In the case that m 2 (p), we have that (m) ⊂ (p) and (m) ⊃ (p), so (m) = (p). In the case that r 2 (p), then we have r = s · p for some element s in the ring R. Substituting, we can see that r = s · (m · r) so 1 = s · m. Then 1 is in the ideal (m), so (m) is necessarily the entire ring. We have thus shown that there do not exist any proper ideals of R containing (p) a prime ideal, and hence it is maximal. 3. Field Extensions This discussion of polynomial rings, maximal ideals, and fields is motivated by the existence of polynomials with no roots in a given field. In general, given a polynomial p(x) in the ring F [x] for any given field F , does there exist a solution to p(x) = 0 that lies in F ? If not, does there exist a field extension of F in which p(x) = 0 has a solution? Consider the classic example: the polynomial x2 + 1 = 0 has no solutions over R. It is easy to see that there are no numbers in R that satisfy this equation - but over the complex numbers, both i and −i are solutions. It is also crucial to notice that the complex numbers, conventionally of the form a + bi, contain a \copy" of the real number line - namely, when b = 0. These concepts are generalized and formalized in this section. Given any poly- nomial with no roots over a given field, we would like to be able to construct a field extension that both contains solutions to the equation and preserves the structure of the original field. Definition 3.1. Let K be a field. A subfield of K is a subset F of K that is closed under the field operations of K. The larger field K is said to be a (field) extension of F . We would like now to use the fact that a polynomial ring F [x] is a principal ideal domain in order to discuss the construction of field extensions. Theorem 3.2. Let R be a ring and I an ideal of R. Then I is maximal if and only if R/I is a field. The proof for this theorem follows from two important results: one is a charac- terization of fields of having only trivial ideals and the other is a correspondence that exists between ideals of a ring R containing an ideal I and ideals of the quotient group R=I. The first result is simply a consequence of the fact that every nonzero 4 JENNY WANG element in a field is a unit. The second result is precisely the Lattice Isomorphism Theorem for rings. Proposition 3.3. A ring R is a field if and only if the only ideals of R are f0g and R. Proof. Suppose that R is a field. Note that f0g is always an ideal of any ring, so f0g is an ideal of R. By definition of a field, every nonzero element of R is a unit, so every non-zero ideal contains a unit. 1 generates the entire ring, so the only nonzero ideal is the entire ring R. Now suppose that the only ideals of R are f0g and R. Let an element a 2 R be nonzero. Then the ideal generated by a is R by assumption: (a) = R.